计算几何

点，线：

poj 1066

题意：给一个矩形，n条在内部线，每次只能通过线段的中点，求从起点走到矩形外要经过几次线段；

显然当我们确定终点后，那些横跨在起点和终点之间的直线是我们无法避免要通过的，且每条直线我们只需要通过一次就好了。

这样我们只要枚举终点，然后判断交点个数就好了。问题的转化，线段平移，或者极端状况。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<queue>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<set>
#define pbk push_back
using namespace std;
const int N=5000+10;
const double eps=1e-10;
inline int dcmp(double x){
    return x<-eps?-1:x>eps;
}
inline double sqr(double x){
    return x*x;
}
struct Point{
    double x,y;
    Point (double x=0,double y=0):x(x),y(y){}
    Point operator - (const Point &p)const{
        return Point(x-p.x, y-p.y );
    }
    Point operator + (const Point &p)const{
        return Point(x+p.x, y+p.y);
    }
    Point operator * (const double &k)const{
        return Point(x*k, y*k);
    }
    Point operator / (const double &k)const{
        return Point(x/k, y/k);
    }
    double operator * (const Point &p)const{
        return x*p.y - y*p.x;
    }
    double operator ^ (const Point &p)const{
        return x*p.x + y*p.y;
    }
    bool operator == (const Point &p)const{
        return dcmp(x-p.x)==0 && dcmp(y-p.y)==0;
    }
    bool operator < (const Point &p)const{
        return dcmp(x-p.x)<0 || (dcmp(x-p.x)==0 && dcmp(y-p.y)<0 );
    }
    double Distance(Point p){
        return sqrt(sqr(x-p.x) + sqr(y-p.y));
    }
    void input(){
        scanf("%lf%lf",&x,&y);
    }
    void ot(){
        printf("%lf %lf\n",x,y);
    }
};
struct Line{
    Point a,b;
    Line(){}
    Line(const Point &_a,const Point &_b):a(_a),b(_b){}
    double operator * (const Point &p)const{
        return (b-a) * (p-a);
    }
    double operator ^ (const Point &p)const{
        return (p-a) ^ (p-b);
    }
    void input(){
        a.input(); b.input();
    }
    void ot(){
        a.ot(); b.ot();
    }
    bool IsPointInLine(const Point &p){
        return !dcmp((*this) * p);
    }
    bool IsPointInSeg(const Point &p){
        return dcmp((*this) *p) == 0 && dcmp( (*this) ^ p ) <= 0;
    }
    bool parallel(Line v){
        return !dcmp( (b-a)*(v.b-v.a) );
    }
    int LineCrossSeg(const Line &v){
        int d1 = dcmp((*this) * v.a), d2 = dcmp((*this) * v.b);
        if ( (d1 ^ d2) == -2) return 2;  // (-1)^(1) == -2;
        return (d1 ==0 || d2 == 0);
    }
    int LineCrossLine(const Line &v){
        if ( (*this).parallel(v) ){
            return dcmp(v*a) == 0;
        }
        return 2;
    }
    Point CrossPoint(const Line &v){
        double s1 = v*a, s2 = v*b;
        return ( a*s2 - b*s1 ) / (s2 - s1);
    }
    int SegCrossSeg(const Line &v){
        int d1 = dcmp( (*this) * v.a);
        int d2 = dcmp( (*this) * v.b);
        int d3 = dcmp(v * a);
        int d4 = dcmp(v * b);
        if ( (d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;// 2 内部相交
        return ( (d1 == 0 && dcmp((*this) ^ v.a) <= 0) // 1 端点相交  0 无交点
              || (d2 == 0 && dcmp((*this) ^ v.b) <= 0)
              || (d3 == 0 && dcmp(v ^ a) <= 0)
              || (d4 == 0 && dcmp(v ^ b) <= 0) );
    }
};
int n;
Line L[31];
vector<Point >p;
Point st;
void solve(){
    int ans = -1;
    for (int i=0;i<p.size();i++){
        Line t = Line(st,p[i]);
        int cnt = 0;
        for (int j=0;j<n;j++){
            if (L[j].SegCrossSeg(t) == 2) cnt++;
        }
        if (cnt<ans || ans==-1) ans=cnt;
    }
    printf("%d\n",ans+1);
}
int main(){
    while (~scanf("%d",&n)){
        printf("Number of doors = ");
    
        p.clear();
        for (int i=0; i<n; i++){
            L[i].input();
            p.pbk(L[i].a); p.pbk(L[i].b);
        }
        st.input();
        if (n<=2){
            printf("1\n"); continue;
        }
        solve();
    }
    
    return 0;
}

 

poj3347

题意：给n个边长为leni的正方形，将它们旋转45度后依次放置，使它们两两不相交且与x轴的交点横坐标最小；

思路：先求出每个正方形与X轴的交点，当前正方形与前面所有正方形相临得到一个xi，最大的xi就是交点；

判断是否可见，对于当前正方形，能挡住它的肯定是边长大于它的正方形，这样我们依次枚举它前面的正方形，求出最大的能延伸的坐标l；

再依次枚举它后面的正方形，求出最小的能延伸的坐标r，如果r>l 就是可见的；   几何，很难？很简单？

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
const double eps = 1e-8;
inline int dcmp(double x){
    return x < -eps ? -1 : x > eps;
}
inline double sqr(double x){
    return x*x;
}
const int N = 50+10;
double x[N],len[N],len2[N];
vector<int > ans;
int n;
void solve(){
    x[0] = len2[0]/2;
    for (int i = 1; i < n; i++) {
        double tmp;
        x[i] = 0;
        for (int j = 0; j < i; j++) {
            if (dcmp(len[i] - len[j]) < 0 ){
                tmp = x[j] + len2[i];
            }else tmp = x[j] + len2[j];
            if (dcmp(tmp - x[i]) > 0) x[i] = tmp;
        }
    }
    ans.clear();
    for (int i = 0; i < n; i++){
        double l = x[i] - len2[i]/2, r = x[i] + len2[i]/2 ;
        for (int j = 0; j < i; j++){
            if (dcmp(len[j] - len[i]) > 0){
                if (dcmp(x[j] + len2[j]/2 - l) > 0) l = x[j] + len2[j]/2;
            }
        }
        for (int j = i+1; j < n; j++){
            if (dcmp(len[j] - len[i]) > 0){
                if (dcmp(x[j] - len2[j]/2 - r) < 0) r = x[j] - len2[j]/2;
            }
        }
        if (dcmp(l - r) < 0) ans.push_back(i+1);  
    }
    int sz = ans.size();
    for (int i = 0; i < sz-1; i++){
        printf("%d ",ans[i]);
    }printf("%d\n",ans[sz-1]);
}
int main(){
    while (~scanf("%d",&n),n){
        for (int i = 0; i < n; i++){
            scanf("%lf",len+i);
            len2[i] = sqrt( len[i]*len[i]*2 );
        }
        solve();
    }    
    return 0;
}

 

poj 2826

题意：两条线段，求能接到多少水。

分析：细节题，一个被坑的状况，上面的线段在交点上方完全覆盖了下面的线段。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#define pbk push_back
using namespace std;
const int N = 300+10;
const double eps = 1e-10;
const int maxP = 10000+10;
inline int dcmp(double x){
    return x<-eps ? -1 : x>eps;
}
inline double sqr(double x){
    return x*x;
}
struct Point {
    double x,y;
    Point(){}
    Point (double a,double b):x(a),y(b){}
    bool operator == (const Point &p)const{
        return (dcmp(x-p.x) == 0 && dcmp(y-p.y) == 0);
    }
    bool operator < (const Point &p)const{
        return dcmp(x-p.x) < 0 || (dcmp(x-p.x) == 0 && dcmp(y-p.y) < 0);
    }
    double operator * (const Point &p)const{
        return x*p.y - y*p.x;
    }
    double operator / (const Point &p)const{
        return x*p.x + y*p.y;
    }

    Point operator - (const Point &p)const{
        return Point(x-p.x,y-p.y);
    }
    Point operator + (const Point &p)const{
        return Point(x+p.x, y+p.y);
    }
    Point operator * (const double &k)const{
        return Point(x*k, y*k);
    }
    Point operator / (const double &k)const{
        return Point(x/k, y/k);
    }
    
    double len2(){
        return x*x + y*y;
    }
    double len(){
        return sqrt(x*x + y*y);
    }
    Point scale(const double &k){//变成长度为k的"向量"
        return dcmp( len() ) ? (*this) * (k / len()) : (*this);
    }
    Point turnLeft(){
        return Point(-y,x);
    }
    Point turnRight(){
        return Point(y,-x);
    }
    void input(){
        scanf("%lf%lf",&x,&y);
    }
    void ot(){
        printf("%lf %lf\n",x,y);
    }
    double Distance(const Point &p){
        return sqrt( sqr(x-p.x) + sqr(y-p.y) );
    }
    Point rotate(const Point &p,double angle ,double k=1){//绕P点k==1逆时针旋转angle； 
        Point vec = (*this) - p;
        double Cos = cos(angle)*k, Sin = sin(angle)*k;
        return p + Point(vec.x * Cos - vec.y * Sin, vec.x * Sin + vec.y * Cos);
    }
};
struct Line{
    Point a,b;
    Line(){}
    Line(Point a,Point b):a(a),b(b){}
    
    double operator * (const Point &p)const{
        return (b-a) * (p-a);
    }
    double operator / (const Point &p)const{
        return (p-a) / (p-b);
    }
    void input(){
        a.input(); b.input();
    }
    void ot(){
        a.ot(); b.ot();
    }
    double len(){
        return a.Distance(b);
    }
    bool parallel(Line v){
        return !dcmp( (b-a)*(v.b-v.a) );
    }
    
    int SegCrossSeg(const Line &v){//2中间相交，1端点相交，0没有交点
        int d1 = dcmp( (*this) * v.a);
        int d2 = dcmp( (*this) * v.b);
        int d3 = dcmp( v * a );
        int d4 = dcmp( v * b );
        if ( (d1^d2) == -2 && (d3^d4) == -2 ) return 2;
        return ( (d1 == 0 && dcmp( (*this)/v.a ) <= 0) 
              || (d2 == 0 && dcmp( (*this)/v.b ) <= 0)
              || (d3 == 0 && dcmp( v / a ) <= 0)
              || (d4 == 0 && dcmp( v / b ) <= 0) );
    }

    int LineCrossSeg(const Line &v){//2相交，1共线，0相离
        int d1 = dcmp( (*this) * v.a), d2 = dcmp( (*this) * v.b);
        if ( (d1^d2) == -2) return 2;
        return (d1 == 0 || d2 == 0);
    }
    int LineCrossLine(const Line &v){//2相交，1共线，0相离
        if ( (*this).parallel(v)){
            return ( dcmp(v * a) == 0);
        }
        return 2;
    }
    Point CrossPoint(const Line &v){
        double s1 = v * a, s2 = v * b;
        return ( a * s2 - b * s1) / ( s2 - s1);
    }
    bool IsPointOnSeg(const Point &p){
        return ( dcmp( (a-p)*(b-p) ) == 0 && dcmp( (p-a)/(p-b) ) <= 0 );
    }
    double DisPointToSeg(Point p){//p点到线段的最短距离 
        if ( a == b) return a.Distance(p);
        Point q = p + (a - b).turnLeft();
        if ( (( p - a )*( q - a )) * ( (p - b)*(q - b) ) > 0 ) {
            return min(p.Distance(a), p.Distance(b));
        }
        return fabs( (*this)*p ) / a.Distance(b) ;
    }
    Point PointToSeg(Point p){//p点到线段最短距离的那个点
        if ( a == b ) return a;
        Point q = p + (a - b).turnLeft();
        if ( (( p - a )*( q - a)) * ((p - b)*(q - b)) > 0){
            return p.Distance(a) < p.Distance(b) ? a : b;
        }
        return CrossPoint( Line(p,q) );
    }
    double DisPointToLine(const Point &p){//点到直线的距离
        return fabs( (*this) * p) / a.Distance(b);
    }
    Point PointToLine(const Point &p){//求点p垂直Line的垂心
        return CrossPoint( Line(p,p + (a - b).turnLeft()) );
    }
    Point SymPoint(const Point &p){//求p关于Line的对称点
        return PointToLine(p) * 2 - p;
    }
    void Move(const double &d){//直线向左侧平移d长度
        Point t = ( b - a ).turnLeft().scale(d);
        a = a + t; b = b + t;
    }
};
struct Polygon{
    int n;
    Point p[maxP];
    void input(){
        for (int i = 0; i < n; i++){
            p[i].input();
        }
    }
    void push(const Point &pp){
        p[n++] = pp;
    }
    void getConvex(Polygon &c){
        sort(p, p+n);
        c.n = n;
        if (n == 0) return;
        c.p[0] = p[0]; if (n == 1) return;
        c.p[1] = p[1]; if (n == 2) return;
        int &top = c.n;
        top =1;
        for (int i = 2; i < n; i++){
            while (top > 0 && dcmp( (c.p[top] - p[i])*(c.p[top-1] - p[i]) ) >= 0)
                top --;
            c.p[++top] = p[i];
        }
        int tmp = top;
        c.p[++top] = p[n-2];
        for (int i = n-3; i >= 0; i--){
            while (top > tmp && dcmp( (c.p[top] - p[i])*(c.p[top-1] - p[i]) ) >= 0)
                top --;
            c.p[++top] = p[i];
        }
    }
    int IsPointInpolygon(const Point &pp){
        p[n] = p[0];
        int wn = 0;
        for (int i =0; i < n; i++) {
            if ( Line(p[i],p[i+1]).IsPointOnSeg(pp) ) {
                if (p[i] == pp || p[i+1] == pp) return 3;
                return 2;
            }
            int k = dcmp( (p[i] - pp)*(p[i+1] - pp) );
            int u = dcmp( p[i].y - pp.y);
            int v = dcmp( p[i+1].y - pp.y);
            if (k > 0 && u < 0 && v >= 0) wn++;
            if (k < 0 && v < 0 && u >= 0) wn--;
        }
        return (wn != 0);
    }
    double CalcPerimeter(){
        if (n == 1) return 0;
        double ret = p[0].Distance(p[n-1]);
        for (int i = 0; i < n-1; i++ ){
            ret += p[i].Distance(p[i+1]);
        }
        return ret;
    }
    double CalcArea(){
        double ret = 0;
        for (int i = 1; i < n-1; i++ ){
            ret += (p[i] - p[0]) * (p[i+1] - p[0]);
        }
        return ret * 0.5;
    }

};
Line L[2];
int main(){
    int T;scanf("%d",&T);
    while (T--){
        L[0].input(); L[1].input();
        if (L[0].parallel(L[1])) {
            printf("0.00\n"); continue;
        }
        if (L[0].a.y<L[0].b.y) swap(L[0].a,L[0].b);
        if (L[1].a.y<L[1].b.y) swap(L[1].a,L[1].b);
        if (L[0].a.y == L[0].b.y || L[1].a.y == L[1].b.y){
            printf("0.00\n"); continue;
        }
        if (L[0].SegCrossSeg(L[1])){
            Point t = L[0].CrossPoint(L[1]);
            if (t == L[0].a || t == L[1].a) printf("0.00\n");
            else {
                int flag = L[0].a.y>L[1].a.y ? 1 : 0;
                Point tmp = L[flag].a;
                if (dcmp((L[flag].a - t)*(L[flag^1].a - t)) <= 0 && dcmp(L[flag^1].a.x - L[flag].a.x) <= 0){
                    printf("0.00\n"); continue;
                }
                if (dcmp((L[flag].a - t)*(L[flag^1].a - t)) >= 0 && dcmp(L[flag^1].a.x - L[flag].a.x) >= 0){
                    printf("0.00\n"); continue;
                }
                Line t2 = Line(tmp,tmp-Point(1,0));
                
                Point tmp2 = t2.CrossPoint(Line(t,L[flag^1].a));
                double ret = fabs((tmp2 - t) * (L[flag].a - t))/2;
                
                printf("%.2lf\n",ret+eps);
            }
        }else printf("0.00\n");
    }
    return 0;
}

 

poj 2074

题意：给你两条平行的线段1,2，然后再给你其他平行的线段Li，问从线段2能看到线段1的最大的连续区间；

分析：枚举线段Li得出对于线段Li在线段2上肯定不能看到线段1的区间，然后顺便把点记录下来

那么最大区间的端点肯定是上面两个区间的端点，然后枚举所有重新组合的区间，取该区间的中点看是否被上面的区间覆盖，

如果没有那么就是有效区间；   要先排除不在线段1,2之间的线段；                  几何，取中点。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#define pbk push_back
using namespace std;
const int N = 300+10;
const double eps = 1e-10;
const int maxP = 10000+10;
inline int dcmp(double x){
    return x<-eps ? -1 : x>eps;
}
inline double sqr(double x){
    return x*x;
}
void in(double &x){
    char c = getchar();
    bool IsN = 0;
    while (c!='-' && (c<'0' || c>'9')) c = getchar();
    if ( c == '-'){
        IsN = 1; x = 0;
    }else x = c - '0';
    c = getchar();
    while (c>='0' && c<='9') {
        x = x*10+c-'0';
        c = getchar();
    }
    if (IsN) x = -x;
}
struct Point {
    double x,y;
    Point(){}
    Point (double a,double b):x(a),y(b){}
    bool operator == (const Point &p)const{
        return (dcmp(x-p.x) == 0 && dcmp(y-p.y) == 0);
    }
    bool operator < (const Point &p)const{
        return dcmp(x-p.x) < 0 || (dcmp(x-p.x) == 0 && dcmp(y-p.y) < 0);
    }
    double operator * (const Point &p)const{
        return x*p.y - y*p.x;
    }
    double operator / (const Point &p)const{
        return x*p.x + y*p.y;
    }

    Point operator - (const Point &p)const{
        return Point(x-p.x,y-p.y);
    }
    Point operator + (const Point &p)const{
        return Point(x+p.x, y+p.y);
    }
    Point operator * (const double &k)const{
        return Point(x*k, y*k);
    }
    Point operator / (const double &k)const{
        return Point(x/k, y/k);
    }
    
    double len2(){
        return x*x + y*y;
    }
    double len(){
        return sqrt(x*x + y*y);
    }
    Point scale(const double &k){//变成长度为k的"向量"
        return dcmp( len() ) ? (*this) * (k / len()) : (*this);
    }
    Point turnLeft(){
        return Point(-y,x);
    }
    Point turnRight(){
        return Point(y,-x);
    }
    void input(){
    //    in(x); in(y);
        scanf("%lf%lf",&x,&y);
    }
    void ot(){
        printf("%lf %lf\n",x,y);
    }
    double Distance(const Point &p){
        return sqrt( sqr(x-p.x) + sqr(y-p.y) );
    }
    Point rotate(const Point &p,double angle ,double k=1){//绕P点k==1逆时针旋转angle； 
        Point vec = (*this) - p;
        double Cos = cos(angle)*k, Sin = sin(angle)*k;
        return p + Point(vec.x * Cos - vec.y * Sin, vec.x * Sin + vec.y * Cos);
    }
};
struct Line{
    Point a,b;
    Line(){}
    Line(Point a,Point b):a(a),b(b){}
    
    double operator * (const Point &p)const{
        return (b-a) * (p-a);
    }
    double operator / (const Point &p)const{
        return (p-a) / (p-b);
    }
    void input(){
        a.input(); b.input();
    }
    void ot(){
        a.ot(); b.ot();
    }
    double len(){
        return a.Distance(b);
    }
    bool parallel(Line v){
        return !dcmp( (b-a)*(v.b-v.a) );
    }
    
    int SegCrossSeg(const Line &v){//2中间相交，1端点相交，0没有交点
        int d1 = dcmp( (*this) * v.a);
        int d2 = dcmp( (*this) * v.b);
        int d3 = dcmp( v * a );
        int d4 = dcmp( v * b );
        if ( (d1^d2) == -2 && (d3^d4) == -2 ) return 2;
        return ( (d1 == 0 && dcmp( (*this)/v.a ) <= 0) 
              || (d2 == 0 && dcmp( (*this)/v.b ) <= 0)
              || (d3 == 0 && dcmp( v / a ) <= 0)
              || (d4 == 0 && dcmp( v / b ) <= 0) );
    }

    int LineCrossSeg(const Line &v){//2相交，1共线，0相离,v线段
        int d1 = dcmp( (*this) * v.a), d2 = dcmp( (*this) * v.b);
        if ( (d1^d2) == -2) return 2;
        return (d1 == 0 || d2 == 0);
    }
    int LineCrossLine(const Line &v){//2相交，1共线，0相离
        if ( (*this).parallel(v)){
            return ( dcmp(v * a) == 0);
        }
        return 2;
    }
    Point CrossPoint(const Line &v){
        double s1 = v * a, s2 = v * b;
        return ( a * s2 - b * s1) / ( s2 - s1);
    }
    bool IsPointOnSeg(const Point &p){
        return ( dcmp( (a-p)*(b-p) ) == 0 && dcmp( (p-a)/(p-b) ) <= 0 );
    }
    double DisPointToSeg(Point p){//p点到线段的最短距离 
        if ( a == b) return a.Distance(p);
        Point q = p + (a - b).turnLeft();
        if ( (( p - a )*( q - a )) * ( (p - b)*(q - b) ) > 0 ) {
            return min(p.Distance(a), p.Distance(b));
        }
        return fabs( (*this)*p ) / a.Distance(b) ;
    }
    Point PointToSeg(Point p){//p点到线段最短距离的那个点
        if ( a == b ) return a;
        Point q = p + (a - b).turnLeft();
        if ( (( p - a )*( q - a)) * ((p - b)*(q - b)) > 0){
            return p.Distance(a) < p.Distance(b) ? a : b;
        }
        return CrossPoint( Line(p,q) );
    }
    double DisPointToLine(const Point &p){//点到直线的距离
        return fabs( (*this) * p) / a.Distance(b);
    }
    Point PointToLine(const Point &p){//求点p垂直Line的垂心
        return CrossPoint( Line(p,p + (a - b).turnLeft()) );
    }
    Point SymPoint(const Point &p){//求p关于Line的对称点
        return PointToLine(p) * 2 - p;
    }
    void Move(const double &d){//直线向左侧平移d长度
        Point t = ( b - a ).turnLeft().scale(d);
        a = a + t; b = b + t;
    }
};
struct Polygon{
    int n;
    Point p[maxP];
    void input(){
        for (int i = 0; i < n; i++){
            p[i].input();
        }
    }
    void push(const Point &pp){
        p[n++] = pp;
    }
    void getConvex(Polygon &c){
        sort(p, p+n);
        c.n = n;
        if (n == 0) return;
        c.p[0] = p[0]; if (n == 1) return;
        c.p[1] = p[1]; if (n == 2) return;
        int &top = c.n;
        top =1;
        for (int i = 2; i < n; i++){
            while (top > 0 && dcmp( (c.p[top] - p[i])*(c.p[top-1] - p[i]) ) >= 0)
                top --;
            c.p[++top] = p[i];
        }
        int tmp = top;
        c.p[++top] = p[n-2];
        for (int i = n-3; i >= 0; i--){
            while (top > tmp && dcmp( (c.p[top] - p[i])*(c.p[top-1] - p[i]) ) >= 0)
                top --;
            c.p[++top] = p[i];
        }
    }
    int IsPointInpolygon(const Point &pp){//包括凹多边形
        p[n] = p[0];
        int wn = 0;
        for (int i =0; i < n; i++) {
            if ( Line(p[i],p[i+1]).IsPointOnSeg(pp) ) {
                if (p[i] == pp || p[i+1] == pp) return 3;
                return 2;
            }
            int k = dcmp( (p[i] - pp)*(p[i+1] - pp) );
            int u = dcmp( p[i].y - pp.y);
            int v = dcmp( p[i+1].y - pp.y);
            if (k > 0 && u < 0 && v >= 0) wn++;
            if (k < 0 && v < 0 && u >= 0) wn--;
        }
        return (wn != 0);
    }
    bool IsPointInConvex(const Point &pp){
        bool s[3] = {false, false, false};
        p[n] = p[0];
        for (int i = 0; i < n; i++){
            s[dcmp(( p[i+1] - pp ) * ( p[i] - pp )) + 1] = true;
            if ( s[0] && s[2] ) return false; //点在外面.
            if ( s[1] ) return true;//点在边上;
        }
        return true;
    }
    double CalcPerimeter(){
        if (n == 1) return 0;
        double ret = p[0].Distance(p[n-1]);
        for (int i = 0; i < n-1; i++ ){
            ret += p[i].Distance(p[i+1]);
        }
        return ret;
    }
    double CalcArea(){
        double ret = 0;
        for (int i = 1; i < n-1; i++ ){
            ret += (p[i] - p[0]) * (p[i+1] - p[0]);
        }
        return ret * 0.5;
    }
    bool IsConvex(){
        bool s[3] = {false, false, false};
        p[n] = p[0], p[n+1] = p[1];
        for (int i = 0; i < n; i++){
            s[dcmp(( p[i+1] - p[i] ) * ( p[i+2] - p[i] )) + 1] = true;
            if (s[0] && s[2]) return false;//如果-1,和1都出现过；
        }
        return true;
    }

}H;
int n;
vector<Point> seg;
vector<double > Lx;
Line t1,t2,L[N];
void solve(){
    double x1,x2,y;
    double ans = 0;
    seg.clear(); Lx.clear();
    Lx.push_back(t2.a.x); Lx.pbk(t2.b.x);

    for (int i = 0; i < n; i++){
        cin>>x1>>x2>>y;
        L[i] = Line(Point(x1,y),Point(x2,y));
    }
    for (int i = 0; i < n; i++){
        if (dcmp(L[i].a.y - t1.a.y) >= 0) continue;
        if (dcmp(L[i].a.y - t2.a.y) < 0) continue;

        Line tp1 = Line(t1.a,L[i].b);
        Line tp2 = Line(t1.b,L[i].a);
        Point p1,p2;
        p1 = tp1.CrossPoint(t2);
        Lx.pbk(p1.x);
        p2 = tp2.CrossPoint(t2);
        Lx.pbk(p2.x);
        seg.pbk(Point(p2.x,p1.x));
    }
    sort(Lx.begin(),Lx.end());
    for (int i = 0; i+1 < Lx.size(); i++){
        if (dcmp(Lx[i] - t2.a.x) < 0) continue;
        if (dcmp(Lx[i+1] - t2.b.x) > 0) break;
        double tx = ( Lx[i] + Lx[i+1] ) / 2;
        int flag = 1;
        for (int j = 0; j < seg.size() && flag; j++){
            if (dcmp(seg[j].x - tx) < 0 && dcmp(seg[j].y - tx) > 0) flag = 0;
        }
        if (flag){
            if (dcmp(Lx[i+1] - Lx[i] - ans) > 0) ans = Lx[i+1] - Lx[i];
        }
    }
    if (dcmp(ans) == 0) printf("No View\n");
    else printf("%.2lf\n",ans);
}
int main(){
    double x1,x2,y;    
    while (cin>>x1>>x2>>y){
        if (dcmp(x1) == 0 && dcmp(x2) == 0 && dcmp(y) == 0) break;
        t1.a = Point(x1,y); t1.b = Point(x2,y);
        cin>>x1>>x2>>y;
        t2.a = Point(x1,y); t2.b = Point(x2,y);
        cin>>n;
        solve();
    }
    return 0;
}

 

 

面：

poj 1228

题意：原本有一个凸包（规则），但是凸包上的一些点不见了，问剩下的点组成的凸包能否确定原先的凸包。

分析：要使剩下的点组成的凸包确定原先的凸包，那么也就是说不管对剩下的凸包在哪里加点都不能形成一个新的面积更大的凸包（剩下的所有点必须在新凸包上），

也就是现在凸包上的每一条线至少有3个点；

如果剩下的点组成的凸包没有面积，或者不能组成凸包，肯定也不行；

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#define pbk push_back
using namespace std;
const int N = 300+10;
const double eps = 1e-10;
const double PI = acos(-1.0);
const int maxP = 10000+10;
inline int dcmp(double x){
    return x<-eps ? -1 : x>eps;
}
inline double sqr(double x){
    return x*x;
}
void in(double &x){
    char c = getchar();
    bool IsN = 0;
    while (c!='-' && (c<'0' || c>'9')) c = getchar();
    if ( c == '-'){
        IsN = 1; x = 0;
    }else x = c - '0';
    c = getchar();
    while (c>='0' && c<='9') {
        x = x*10+c-'0';
        c = getchar();
    }
    if (IsN) x = -x;
}
struct Point {
    double x,y;
    Point(){}
    Point (double a,double b):x(a),y(b){}
    bool operator == (const Point &p)const{
        return (dcmp(x-p.x) == 0 && dcmp(y-p.y) == 0);
    }
    bool operator < (const Point &p)const{
        return dcmp(x-p.x) < 0 || (dcmp(x-p.x) == 0 && dcmp(y-p.y) < 0);
    }
    double operator * (const Point &p)const{
        return x*p.y - y*p.x;
    }
    double operator / (const Point &p)const{
        return x*p.x + y*p.y;
    }

    Point operator - (const Point &p)const{
        return Point(x-p.x,y-p.y);
    }
    Point operator + (const Point &p)const{
        return Point(x+p.x, y+p.y);
    }
    Point operator * (const double &k)const{
        return Point(x*k, y*k);
    }
    Point operator / (const double &k)const{
        return Point(x/k, y/k);
    }
    
    double len2(){
        return x*x + y*y;
    }
    double len(){
        return sqrt(x*x + y*y);
    }
    Point scale(const double &k){//变成长度为k的"向量"
        return dcmp( len() ) ? (*this) * (k / len()) : (*this);
    }
    Point turnLeft(){
        return Point(-y,x);
    }
    Point turnRight(){
        return Point(y,-x);
    }
    void input(){
    //    in(x); in(y);
        scanf("%lf%lf",&x,&y);
    }
    void ot(){
        printf("%lf %lf\n",x,y);
    }
    double Distance(const Point &p){
        return sqrt( sqr(x-p.x) + sqr(y-p.y) );
    }
    Point rotate(const Point &p,double angle ,double k=1){//绕P点k==1逆时针旋转angle； 
        Point vec = (*this) - p;
        double Cos = cos(angle)*k, Sin = sin(angle)*k;
        return p + Point(vec.x * Cos - vec.y * Sin, vec.x * Sin + vec.y * Cos);
    }
};
struct Line{
    Point a,b;
    Line(){}
    Line(Point a,Point b):a(a),b(b){}
    
    double operator * (const Point &p)const{
        return (b-a) * (p-a);
    }
    double operator / (const Point &p)const{
        return (p-a) / (p-b);
    }
    void input(){
        a.input(); b.input();
    }
    void ot(){
        a.ot(); b.ot();
    }
    double len(){
        return a.Distance(b);
    }
    bool parallel(Line v){
        return !dcmp( (b-a)*(v.b-v.a) );
    }
    
    int SegCrossSeg(const Line &v){//2中间相交，1端点相交，0没有交点
        int d1 = dcmp( (*this) * v.a);
        int d2 = dcmp( (*this) * v.b);
        int d3 = dcmp( v * a );
        int d4 = dcmp( v * b );
        if ( (d1^d2) == -2 && (d3^d4) == -2 ) return 2;
        return ( (d1 == 0 && dcmp( (*this)/v.a ) <= 0) 
              || (d2 == 0 && dcmp( (*this)/v.b ) <= 0)
              || (d3 == 0 && dcmp( v / a ) <= 0)
              || (d4 == 0 && dcmp( v / b ) <= 0) );
    }

    int LineCrossSeg(const Line &v){//2相交，1共线，0相离,v线段
        int d1 = dcmp( (*this) * v.a), d2 = dcmp( (*this) * v.b);
        if ( (d1^d2) == -2) return 2;
        return (d1 == 0 || d2 == 0);
    }
    int LineCrossLine(const Line &v){//2相交，1共线，0相离
        if ( (*this).parallel(v)){
            return ( dcmp(v * a) == 0);
        }
        return 2;
    }
    Point CrossPoint(const Line &v){
        double s1 = v * a, s2 = v * b;
        return ( a * s2 - b * s1) / ( s2 - s1);
    }
    bool IsPointOnSeg(const Point &p){
        return ( dcmp( (a-p) * (b-p) ) == 0 && dcmp( (p-a)/(p-b) ) <= 0 );
    }
    double DisPointToSeg(Point p){//p点到线段的最短距离 
        if ( a == b) return a.Distance(p);
        Point q = p + (a - b).turnLeft();
        if ( (( p - a )*( q - a )) * ( (p - b)*(q - b) ) > 0 ) {
            return min(p.Distance(a), p.Distance(b));
        }
        return fabs( (*this)*p ) / a.Distance(b) ;
    }
    Point PointToSeg(Point p){//p点到线段最短距离的那个点
        if ( a == b ) return a;
        Point q = p + (a - b).turnLeft();
        if ( (( p - a )*( q - a)) * ((p - b)*(q - b)) > 0){
            return p.Distance(a) < p.Distance(b) ? a : b;
        }
        return CrossPoint( Line(p,q) );
    }
    double DisPointToLine(const Point &p){//点到直线的距离
        return fabs( (*this) * p) / a.Distance(b);
    }
    Point PointToLine(const Point &p){//求点p垂直Line的垂心
        return CrossPoint( Line(p,p + (a - b).turnLeft()) );
    }
    Point SymPoint(const Point &p){//求p关于Line的对称点
        return PointToLine(p) * 2 - p;
    }
    void Move(const double &d){//直线向左侧平移d长度
        Point t = ( b - a ).turnLeft().scale(d);
        a = a + t; b = b + t;
    }
};
struct Polygon{
    int n;
    Point p[maxP];
    void input(){
        for (int i = 0; i < n; i++){
            p[i].input();
        }
    }
    void push(const Point &pp){
        p[n++] = pp;
    }

    void getConvex(Polygon &c){
        sort(p, p+n);
        c.n = n;
        if (n == 0) return;
        c.p[0] = p[0]; if (n == 1) return;
        c.p[1] = p[1]; if (n == 2) return;
        int &top = c.n;
        top =1;
        for (int i = 2; i < n; i++){
            while (top > 0 && dcmp( (c.p[top] - p[i])*(c.p[top-1] - p[i]) ) >= 0)
                top --;
            c.p[++top] = p[i];
        }
        int tmp = top;
        c.p[++top] = p[n-2];
        for (int i = n-3; i >= 0; i--){
            while (top > tmp && dcmp( (c.p[top] - p[i])*(c.p[top-1] - p[i]) ) >= 0)
                top --;
            c.p[++top] = p[i];
        }
    }
    int IsPointInpolygon(const Point &pp){//包括凹多边形
        p[n] = p[0];
        int wn = 0;
        for (int i =0; i < n; i++) {
            if ( Line(p[i],p[i+1]).IsPointOnSeg(pp) ) {
                if (p[i] == pp || p[i+1] == pp) return 3;
                return 2;
            }
            int k = dcmp( (p[i] - pp)*(p[i+1] - pp) );
            int u = dcmp( p[i].y - pp.y);
            int v = dcmp( p[i+1].y - pp.y);
            if (k > 0 && u < 0 && v >= 0) wn++;
            if (k < 0 && v < 0 && u >= 0) wn--;
        }
        return (wn != 0);
    }
    bool IsPointInConvex(const Point &pp){
        bool s[3] = {false, false, false};
        p[n] = p[0];
        for (int i = 0; i < n; i++){
            s[dcmp(( p[i+1] - pp ) * ( p[i] - pp )) + 1] = true;
            if ( s[0] && s[2] ) return false; //点在外面.
            if ( s[1] ) return true;//点在边上;
        }
        return true;
    }
    double CalcPerimeter(){
        if (n == 1) return 0;
        double ret = p[0].Distance(p[n-1]);
        for (int i = 0; i < n-1; i++ ){
            ret += p[i].Distance(p[i+1]);
        }
        return ret;
    }
    double CalcArea(){
        double ret = 0;
        for (int i = 1; i < n-1; i++ ){
            ret += (p[i] - p[0]) * (p[i+1] - p[0]);
        }
        return ret * 0.5;
    }
    bool IsConvex(){
        bool s[3] = {false, false, false};
        p[n] = p[0], p[n+1] = p[1];
        for (int i = 0; i < n; i++){
            s[dcmp(( p[i+1] - p[i] ) * ( p[i+2] - p[i] )) + 1] = true;
            if (s[0] && s[2]) return false;//如果-1,和1都出现过；
        }
        return true;
    }

}H,H2;
int n;
int check(Line L){
    for (int i = 0; i < H.n; i++ ){
        if (L.IsPointOnSeg(H.p[i]) && !(H.p[i] == L.a) && !(H.p[i] == L.b)) return 1;
    }
    return 0;
}
void solve(){
    H.getConvex(H2);
    H2.p[H2.n] = H2.p[0];
    int flag = 1;
    if (dcmp( H2.CalcArea() ) <= 0) flag = 0;
    for (int i = 0; i < H2.n && flag; i++ ){
        if ( check( Line(H2.p[i],H2.p[i+1]) ) == 0 ) flag = 0;
    }
    puts(flag ? "YES":"NO");

}
int main(){
    int T; scanf("%d",&T);
    while (T--){
        scanf("%d",&H.n);
        H.input();
        solve();
    }
    return 0;
}